Question

1 A converging lens with a focal length of 12.2 cm forms a virtual image 7.9mm tall, 11 2emto right of the lens. a. Determine the position of the object. b. Determine the size of the object. Is the image upright or inverted? Are the object and image on the same side or opposite sides of the lens? c. d. 2 You want to use a lens with a focal length of magnitude 36cm with the image twice as long as the object itself. to produce a real image of an object, What kind of lens do you need? Where should the object be placed? Suppose you want a virtual image of the same object, with the same magnitude magnification- what kind of lens do you need? Where should the object be placed if the magnitude of the focal length is 36cm? a. b. c. d. 3. A converging lens with a focal length of 90.5 cm forms an image of a 3.25 cm-tall real object that is the left of the lens. The image is 4.7cm tall and inverted. Where are the object and image located in relation to the lens? Is the image real or virtual? a. b.

Answer 1

11.

f = 12.2 cm

v = 17.2 cm

a)

using lens formula

1/f = 1/ v - 1/u

1/ 12.2 = 1/17.2 + 1/ u

u = 41.97 cm

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b)

as we know,

m = v / u = h'/h

h' = v h / u

h' = 17.2 * 7.9 / 41.97 = 3.238 mm

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c)

since the image formed right to the lens so it must be real and inverted.

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d)

object and image are opposite side of the lens

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